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How do you prove $|\sqrt[3]{x} - \sqrt[3]{y}|$ $\leq$ $\sqrt[3]{|x-y|}$?

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I corrected your TeX so that the minus sign was inside dollar signs. – Mariano Suárez-Alvarez♦ Oct 13 '10 at 2:39
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This is false. $x=8$, $y=-1$. – Aryabhata Oct 13 '10 at 2:49
Do you require the cube roots to be real as well, i.e. x,y \geq 0? – WWright Oct 13 '10 at 2:52
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Following up on Moron's comment, I guess you are missing the condition that $x$ and $y$ have the same sign (in which case you may as well assume that they are both positive). In this case the inequality is true, and the now-deleted comment gave one good method of solution: cube both sides, and compare them to get the desired inequality. (It will help to assume that $x > y$, as you may (otherwise switch them, and nothing changes). Also, for psychological purposes, it may help to write $x^{1/3} = a$ and $y^{1/3} = b$, as the deleted comment suggested.) – Matt E Oct 13 '10 at 2:55

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It's not true. Try $x=1$ and $y=-1$. If you assume $x$ and $y$ have the same sign, and you might as well assume $x\gt y\gt 0$, then it reduces to showing $(x-y)^3\leq x^3-y^3$ (WLOG replacing variables in the original inequality with cubes and cubing both sides). This is true because $x\gt y\gt 0$ implies $3xy^2<3x^2y$.

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Moron's and Matt E's comments were posted within a few minutes of my answer. Moron noticed what happens when $x$ and $y$ have opposite signs before I posted, and Matt E gave a better written explanation for the same sign case. – Jonas Meyer Oct 13 '10 at 3:16

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